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Simple Velocity question

  1. Jun 17, 2008 #1
    1. The problem statement, all variables and given/known data

    The instantaneous velocity of a parrticle is given by the expression:
    [tex]v=u sin \omega t [/tex]
    The time of complete cycle is [tex]T= \frac{2 \pi}{\omega} [/tex]

    Find out the average speed!


    3. The attempt at a solution
    After integration;
    i get the following answer:
    [tex] \frac{2u} { \pi}[/tex]

    tell me if I am wrong I will show my working if I am wrong.
    Thanks a lot!
     
  2. jcsd
  3. Jun 17, 2008 #2
    Hi ritwik

    The average speed over the time of a complete cycle? I think you made a mistake. Look at the sin function again and think of it's average value. It can be done without integrating or any calculus whatsoever. Or am I wrong?
     
  4. Jun 17, 2008 #3
    Lol!! Thats so clever armis. But whatever your method might be, the answer will be same. So what did u get?
    Dont help people unless u get the answer urself. It might confuse the beginners like me all the more.
     
  5. Jun 17, 2008 #4

    alphysicist

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    Hi ritwik06,

    Your result looks right to me.
     
  6. Jun 17, 2008 #5
    ritwik, your result is wrong.
    what limits did you take for integrating?
     
  7. Jun 17, 2008 #6
    Well, I only wanted to help and solve the problem in the process. So were is no reason to laugh at me

    I just thought that if velocity varies from u to -u then the average value will be just zero. Of course if we are talking about the absolute value of u then [tex] \frac{2u}{\pi} [/tex] is the answer

    Sorry, I didn't mean that
     
  8. Jun 17, 2008 #7

    alphysicist

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    Hi jablonsky27,

    I got the same answer as ritwik. Are you integrating the correct function (they ask for the average speed)?
     
  9. Jun 17, 2008 #8
    hi alphysicist,
    you dont even have to integrate the given function. the average value of a sine(or cosine) over one cycle is always zero. the velocity function mentioned, is a sine with no additional constant term. so its average value too over one cycle will be zero.
    if, you really want to calculate it,


    Vaverage = (1/T).[tex]\int[/tex](u.sinwt) over the limits 0 to T,

    = (u/T).[-cos(w.2/[tex]\Pi[/tex]T) + cos(0)]

    = (u/T).[-1 + 1]

    = 0

    hopefully, it makes sense. first time with Latex. :biggrin:
     
  10. Jun 17, 2008 #9
    Well if you go back and read the question, its written in bold face:
    average speed
     
  11. Jun 17, 2008 #10
    Hey, I wasnt laughing at you dear. I was just wonering how u could take that out without integration. I mean the average speed. Anyways, thanks a lot for the effort.
     
  12. Jun 17, 2008 #11

    alphysicist

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    jablonsky,

    I see ritwik already remarked about how we are looking for the average speed (not average velocity). So the integral to calculate would be:




    [tex]
    \mbox{average speed }= \frac{1}{T}\ \int\limits_0^Tdt\ |u \sin(\omega t)|
    [/tex]

    Or you could restrict the integral to just one-fourth of a period.
     
  13. Jun 17, 2008 #12
    oh ya. sorry. should learn to read questions more carefully in the future.
     
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